For 0≤r<2n, 2n+rCn 2n−rCn cannot exceed

For 0r<2n, 2n+rCn 2nrCn cannot exceed

  1. A

     4nCn

  2. B

     4nC2n

  3. C

     6nC3n

  4. D

    none of these.

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    Solution:

    For 0r<2n,

    (1+x)4n=(1+x)2n+r(1+x)2nr=A0+A1x+A2x2++A2n+rx2n+rB0+B1x+B2x2++B2nrx2nr 

    where Ak=2n+rCk(0k2n+r)

    and Bk=2nrCk(0k2nr)

    Coefficient of x2n on the RHS

    A2nB0+A2n1B1++AnBn+An+1Bn1++ArB2nr         

    = coefficient of x2n on LHS

    =4nC2n.

    Thus, AnBn<4nC2n 

     2n+rCn 2nrCn<4nC2n

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